Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 - 2}{x - 5} = \dfrac{8x - 17}{x - 5}$
Multiply both sides by $x - 5$ $ \dfrac{x^2 - 2}{x - 5} (x - 5) = \dfrac{8x - 17}{x - 5} (x - 5)$ $ x^2 - 2 = 8x - 17$ Subtract $8x - 17$ from both sides: $ x^2 - 2 - (8x - 17) = 8x - 17 - (8x - 17)$ $ x^2 - 2 - 8x + 17 = 0$ $ x^2 + 15 - 8x = 0$ Factor the expression: $ (x - 3)(x - 5) = 0$ Therefore $x = 3$ or $x = 5$ However, the original expression is undefined when $x = 5$. Therefore, the only solution is $x = 3$.